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Problem

I want to know how to solve the differential equation $$ \dot{x} + a\cdot x - b\cdot \sqrt{x} = 0 $$ for $a>0$ and both situations: for $b > 0$ and $b < 0$.

My work

One can separate the variables to obtain: $$ \frac{dx}{b\cdot \sqrt{x} - a\cdot x} = dt$$ but I do not know how to proceed ... https://www.wolframalpha.com/input/?i=solve+x%27(t)%2Bax(t)-bsqrt(x(t))+%3D+0 it seems to have an explicit solution ...

Context

This problem occurs in the following context: $$ \ddot{X} + a \cdot \dot{X} = f(X)$$ then multiplying both sides with $2\dot{X}^T$ one obtains: $$ (\dot{X}^T\dot{X})' + 2a\cdot \dot{X}^T\dot{X} = 2\dot{X}^T f(X)$$ Let $v= \dot{X}^T \dot{X}$ and the above differential equation arises ...

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    $\begingroup$ It seems that this is a Bernoulli equation ... $\endgroup$ – C Marius May 14 at 19:02
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The next step is to integrate,

$$t+c=\int\frac{dx}{b\sqrt x-ax}=\int\frac{d\sqrt x^2}{b\sqrt x-a\sqrt x^2}=\int\frac{2\,d\sqrt x}{b-a\sqrt x}=-\frac2a\log\left(\frac ba-\sqrt x\right).$$

From this you can draw $x$,

$$x=\left(\frac ba-e^{-a(t+c)/2}\right)^2.$$

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  • $\begingroup$ ...along with the trivial solution $x = 0$ (which was lost when OP divided by something with $\sqrt{x}$ as a factor). $\endgroup$ – John Hughes May 14 at 19:57
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Some caution has to be taken here because of the presence of the $\sqrt{}$ function--we need to keep track of signs carefully. We know that $x(t) \ge 0$ everywhere, so there is a function $u$ such that $u \ge 0$ and $x = u^2$. Additionally, we can scale $u$ and $t$ to eliminate the constants. The proper choice for this is $x(t) = (b/a)^2 u(a t/2)^2$, giving $$ u\, [u'+u - \mathrm{sgn}(b)] = 0 $$ which has two solutions: $u(s)= 0$ and $u(s) = \mathrm{sgn}(b)+Ce^{-s}$. For this second solution, $x(t) = (b/a)^2 [\mathrm{sgn}(b)+Ce^{-s}]^2$. Factoring $|b|/a$ into the brackets gives $$ x(t) = \left(\frac{b}{a} + Ce^{-at/2}\right)^2 \;\;\;\;\mathrm{or}\;\;\; x(t) = 0 $$ Except, there's one little problem here. We required $u\ge 0$, but that term in parentheses, which has the same sign as $u$, could be negative. The key is to note that for both solutions, when $x(t) = 0$, $x'(t)$ is also $0$. This allows them to be spliced together at that point and still be continuously differentiable. Thus, the general solution is $$ x(t) = \left[\max \left(\frac{b}{a} + Ce^{-at/2},0\right)\right]^2 $$ for an arbitrary real constant $C$.

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One way of working this out is to make the substitution $y = \sqrt{x}$. Then, $$\frac{dx}{b\sqrt{x}-ax} \rightarrow \frac{2ydy}{by-ay^2} = \frac{2ydy}{y(b-ay)}.$$ You can treat the integral in $y$ with partial fractions.

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    $\begingroup$ Much faster to simplify $\dfrac2{b-ay}dy$ ! $\endgroup$ – Yves Daoust May 14 at 19:20

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